implicit differentiation computation

Question

I am having troubles isolating and solving for $\frac{dy}{dx}$,

$\sin(x+y)=xy$

$\cos(x+y)(1+\frac{dy}{dx})= x\frac{dy}{dx} + y$

What would be the next few step to solve for $\frac{dy}{dx}$?

Answer 1

$$ \cos(x+y)\left(1+ \frac{dy}{dx}\right)= x\frac{dy}{dx} + y \qquad \longleftarrow \text{ error: You need $y$ here, not $x$.} $$ Expand: $$ \cos(x+y)\cdot1 + \cos(x+y)\frac{dy}{dx} = x\frac{dy}{dx} + y $$ Transpose: $$ \cos(x+y)\frac{dy}{dx} - x\frac{dy}{dx} = \cos(x+y)\cdot1 + y $$ Factor $$ (\cos(x+y) - x) \frac{dy}{dx} = \cos(x+y)\cdot1 + y $$ Then divide both sides by $\cos(x+y) - x$.

Answer 2

Hoping that you already know it, the implicit function theorem is very useful.

In you case, consider the function $$F=\sin(x+y)-xy=0$$ Computing each derivative, considering that the other variable is a constant, gives $$F'_x=\cos(x+y)-y$$ $$F'_y=\cos(x+y)-x$$ $$y'=\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{\cos(x+y)-y}{\cos(x+y)-x}$$